Question

Verify that
$y = a \: cosx + b \: sinx$
is a solution of the differential equation:
$d {}^{2}y \div dy {}^{2} + y = 0$
​

Answer 1

y=a cos x+b sin ;

(dy)/(dx) = derivative of a * cos x + b * sin x for x

=a d(cos x) dx +b d(sin x) dx

=a(-sin x)+b(cos x)

=-a sinx+b cosx

(d ^ 2 * y)/(d * x ^ 2) = derivative of (dy)/(dx) for x

= d dx (-a sin x+b cosx)

=-a d(sin x) dx +b d(cos x) dx

=-a(cos x)+b(-sin x)

=-a cos x-b sin x

Now, we have to verify (d ^ 2 * y)/(d * x ^ 2) + y = 0

Taking L.H.S

(d ^ 2 * y)/(d * x ^ 2) + y = (- a * cos x - b * sin x) + (a * cos x + b * sin x)

=0.

=RHS.

HENCE VERIFIED.